Ramanujan's series $1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}$

Question

Ramanujan gave the following series evaluation $$1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\sqrt{\pi}\Gamma^{2}\left(\dfrac{3}{4}\right)}$$ in his first and famous letter to G H Hardy. The form of the series is similar to his famous series for $1/\pi$ and hence a similar approach might work to establish the above evaluation. Thus if $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ then Ramanujan's series is equal to $f(1)+8f'(1)$. Unfortunately the series for $f(x) $ does not appear to be directly related to elliptic integrals or amenable to Clausen's formula used in the proofs for his series for $1/\pi$.

Is there any way to proceed with my approach? Any other approaches based on hypergeometric functions and their transformation are also welcome.

Update: This question is now solved, thanks to the people at mathoverflow.

Answer 1

Considering $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ $$\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ and, thanks to a CAS, $$f(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ By the way,

$$g(x)=1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^n$$ write $$g(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)+\frac{x} {32} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$

Edit

Just out of curioisity, considering $$a_n=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ I had a look at functions $$f_k(x)=1+\sum_{n=1}^{\infty} a_n^k\, x^n$$ and their derivatives and found (probably trivial) the following $$f_2(x)=\, _2F_1\left(\frac{1}{4},\frac{1}{4};1;x\right)$$ $$f_3(x)=\, _3F_2\left(\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1;x\right)$$ $$f_4(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ $$f_5(x)=\, _5F_4\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1,1;x \right)$$ and so on. Similarly $$f_2'(x)=\frac{1}{16} \, _2F_1\left(\frac{5}{4},\frac{5}{4};2;x\right)$$ $$f_3'(x)=\frac{1}{64} \, _3F_2\left(\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2;x\right)$$ $$f_4'(x)=\frac{1}{256} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$ $$f_5'(x)=\frac{1}{1024}\, _5F_4\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2,2;x \right)$$

For $x=1$ $$f_2(1)=\frac{\Gamma \left(\frac{1}{4}\right)}{\sqrt{2 \pi } \Gamma \left(\frac{3}{4}\right)}$$ $$f_3(1)=\frac{\sqrt{\pi }}{\sqrt[4]{2} \Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{7}{8}\right)^2}$$ but, for sure, I have not be able to identify the next terms.

Answer 2

(Too long for a comment. But these factoids might be useful.) If I remember correctly, a pair of series in that letter (plus an additional one) was,

\begin{aligned} U_1 &= 1-5\left(\frac{1}{2}\right)^{3}+\,9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}\,-\,13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}\,+\,\cdots=\dfrac{2}{\Big[\Gamma\big(\tfrac{1}{2}\big)\Big]^2}=\dfrac{2}{\pi} \\[4pt] V_1 &= 1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\Big[\Gamma\big(\tfrac{3}{4}\big)\Big]^2\sqrt{\pi}}\\[4pt] W_1 &= 1-5\left(\frac{1}{2}\right)^{5}+\,9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{5}\,-\,13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{5}\,+\,\cdots=\dfrac{2}{\Big[\Gamma\big(\tfrac{3}{4}\big)\Big]^4} \end{aligned}

Their similar form can be enhanced by gamma functions as,

\begin{aligned}U_1&=\sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\V_1&=\sum_{n=0}^\infty (8n+1)\left(\frac{\Gamma\big(n+\tfrac{1}{4}\big)}{n!\;\Gamma\big(\tfrac{1}{4}\big)}\right)^4\end{aligned}

$U_1$ belongs to an infinite family,

\begin{align} U_1 &=\frac{4}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 (-1)^n \color{blue}{\frac{4n+1}{\;(2^6)^{n+1/2}}}\\ U_2 &=\frac{4}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 (-1)^n \color{blue}{\frac{12n+2}{\;(2^9)^{n+1/2}}}\\ U_3 &=\frac{4}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{6n+1}{\;(2^8)^{n+1/2}}}\\ U_4 &=\frac{4}{\pi}=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{42n+5}{\;(2^{12})^{n+1/2}}} \end{align}

and so on. The Chudnovsky formulas are level 1, but these are the level 4 versions. And while the former uses the j-function, the latter uses the function,

$$r_4(\tau) = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)}\right)^{24}$$

with the Dedekind eta function $\eta(\tau)$. The four above uses $\tau =\frac{1+\sqrt{-2}}2,\,\frac{1+\sqrt{-4}}2,\,\frac{\sqrt{-3}}2,\,\frac{\sqrt{-7}}2$ and are the only values such that $r_4(\tau) = -2^6,\,-2^9,\,2^8,\,2^{12}$ is an algebraic number of degree 1 (an integer) and the formula converges. (Other values yield algebraic numbers of degree 2,3, etc., so there are infinitely many convergent formulas.)

So $V_1$ may then also belong to an infinite family.

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To Jack: As relations were asked, maybe the one below will help? Given the binomial $\binom nk$, then we have,

$$\binom{-\tfrac14}{n}\binom{-\tfrac34}{n} = \binom{-\tfrac34}{n}\frac{(-1)^n\,\Gamma\big(n+\tfrac{1}{4}\big)}{\Gamma\big(\tfrac{1}{4}\big)\Gamma(n+1)} = \frac{\binom{4n}{2n}\binom{2n}{n}}{64^n}$$